Odpowiedź :
Odpowiedź:
zad 1
S = (- 3 , 2 ) , B = ( 1 , - 6 ) , A = (xa , ya)
xs = - 3 , xb = 1 , ys = 2 , yb = - 6
xs = (xa + xb)/2
2 * xs = xa + xb
xa = 2 * xs - xb = 2 * (- 3) - 1 = - 6 - 1 = - 7
ys = (ya + yb)/2
2 * ys = ya + yb
ya = 2 * ys - yb = 2 * 2 - (- 6) = 4 + 6 = 10
A = (- 7 , 10 )
IABI = √[(xb - xa)² + (yb - ya)²] = √[(1 + 7)² + (- 6 - 10)²] = √{8² + (- 16)²] =
= √(64 + 256) = √320 = √(64 * 5) = 8√5 [j]
[j] - znaczy właściwa jednostka
zad 2
A = (xa , 8 ) , B = ( - 4 , yb) , S = (1 , 4 )
xs = (xa + xb)/2
2 * xs = xa + xb
xa = 2 * xs - xb = 2 * 1 - (- 4) = 2 + 4 = 6
ys = (ya + yb)/2
2 * ys = ya + yb
yb = 2 * ys - ya = 2 * 4 - 8 = 8 - 8 = 0
A = ( 6 , 8 ) , B = ( - 4 , 0 )
zad 3
(x - 1)² + (y + 2)² = 9
S₁ - współrzędne środka okręgu = (1 , - 2)
x² + y² + 6x - 10y - 1 = 0
x² + y² - 2 * (- 3)x - 2 * 5y - 1 = 0
S₂ - środek drugiego okręgu = ( - 3 , 5 )
IS₁S₂I = √[(- 3 - 1)² + ( 5 + 2)²] = √[(- 4)² + 7²] = √(16 + 49) = √65 [j]